Chapter 15, Problem 12RP

To determine

To find: The minimum integral value in the given triangle.

Answer to Problem 12RP

  $\text{Integral value of }x=5.$

Explanation of Solution

Given:

  $\begin{array}{l}\text{In }\Delta ABC\\ AB>BC\\ \angle C=4x-4\\ \angle A=x+9\end{array}$

Formula Used:

In a triangle angle opposite to larger side is always larger, and smallest side is smallest.

  $\begin{array}{l}If\\ a>b\\ b>c\\ then\\ a>c\end{array}$

Calculation:

  

  $\begin{array}{l}AB>AC\\ \text{⇒}4x-4>x+9\\ \text{⇒}3x>13\\ \text{⇒}x>4.33\\ so\text{ minimum integral value of }x=5.\end{array}$

Conclusion:

Thus the $\text{Minimum integral value of }x=5$